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You are able to begin the M.Tech admission process by going to the official website of Jhulelal Institute of Technology. You register online, submit the forms, upload the necessary documents and pay the cost of applying. Applying for a course is easy and done entirely online. Check that you satisfy the requirements before you apply.

To be eligible for M.Tech at Jhulelal Institute of Technology, applicants must have studied for a bachelor's degree and gained 50% marks. Getting a valid score on GATE or GPAT is essential for choosing candidates. Both grades and entrance exam marks are taken into account by the institute. You need both for admission to the school.

35. Let, z = a + ib

= $\frac{{\left(x+i\right)}^2}{2x^2+ 1}$

= $\frac{x^2+ i^2+2xi}{2x^2+ 1}$ [since, (a + b)² = a² + b² + 2ab]

= $\frac{x^2- 1+2xi}{2x^2+ 1}$ [since, i2 = –1]

= $\frac{x^2- 1}{2x^2+ 1}$ + $\frac{i2x}{2x^2+ 1}$

So, |z|² = a² + b²

= $\frac{{\left(x^2- 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ + $\frac{{(2x)}^2}{{\left(2x^2+ 1\right)}^2}$

= $\frac{{\left(x^2\right)}^2+ 1^2- 2.x^2.1+ 4x^2}{{\left(2x^2+ 1 \right)}^2}$ [since, (a + b)² = a² + b² + 2ab]

= $\frac{x^4+ 1-2x^2+ 4x^2}{{\left(2x^2+ 1\right)}^2}$

= $\frac{x^4+ 2x^2+ 1}{{\left(2x^2+ 1\right)}^2}$

= $\frac{{\left(x^2+ 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ [as, (a + b)² = a² + b² + 2ab]

Hence proved.

Hi, Listed below are the top BA colleges in Mumbai along with their average placement packages and the top recruiters:

Disclaimer: This information is sourced from official website/ media reports/ NIRF website and may vary.

BTech admission to Jhulelal Institute of Technology is open to students using JEE Main and MHT CET as valid admission scores. Students who want to apply should have completed Class 12 and studied Physics, Chemistry and Mathematics. You must achieve the required score to be picked for the course.

Students applying for BTech at Jhulelal Institute of Technology need to use the online application form available on the official site. Registering, filling in the form, uploading files and paying the fee are all part of the process. Students are admitted depending on their Class 12 scores and marks in JEE Main or MHT CET. It has been set up so students can find things easier.

34. z₁ = 2 – i ,z₂ = 1 + i

$\left|\frac{z_1+ z_2+ 1}{z_1- z_2+ 1}\right|$

= $\left|\frac{2-i+1+i+1}{2-i-1-i+1}\right|$

= $\left|\frac{4}{2-2i}\right|$

= $\left|\frac{4}{2\left(1-i\right)}\right|$

= $\left|\frac{2}{1i}\right|$

= $|\frac{2}{1-i}$ × $\frac{1+i}{1+i}|$ [multiply numerator and denominator by (1 + i)]

= $\left|\frac{2+2i}{1^2-i^2}\right|$

= $\left|\frac{2+2i}{1-\left(-1\right)}\right|$ [since, i2 = –1]

= $\left|\frac{2+2i}{1+1}\right|$

= $\left|\frac{2\left(1+i\right)}{2}\right|$

= |1 + i|

Yes, joining te top BA colleges in Mumbai (All) can be worthwhile for you. For that, you need to check its ROI. Listed below are some colleges along with their tuition fees and average placement package:

Disclaimer: This information is sourced from official website/ media reports/ NIRF website and may vary.

All parts of the Jhulelal Institute of Technology application process happen online and can be followed easily. To apply, individuals just have to go to the main site, enrol and fill in the form. Once you have sent the required documentation, you pay the upfront fee. Aiming to save time, the process has been made as simple as possible for all students.

For applying to Jhulelal Institute of Technology, follow the link to the official site. Click the Apply Now button and then use your basic information to sign up and get login details. After you have created your account, finish the form, submit your documents and make the payment. When your application is submitted, it will be looked over for further processing.

Yes, there are many affordable BA colleges in Mumbai (All). Some of them are mentioned below along with their tuition fees of less than INR 1 lakh:

Disclaimer: This information is sourced from official website.

32. 27x² – 10x + 1 = 0

Comparing the given equation with ax² + bx + c = 0

We have, a = 27, b = –10  andc = 1

Hence, discriminant of the equation is

b² – 4ac = ( 10)² – 4 * 27 * 1 = 100 – 108 =  –8

Therefore, the solution of the quadratic equation is

31.

Multiplying the above equation by 2, we get

and Comparing with ax² + bx + c = 0

We have, a = 2, b = –4 and c = 3

Hence, discriminant of the equation is

b² – 4ac = (-4)² – 4 * 2* 3 = 16 – 24 = –8

Therefore, the solution of the quadratic equation is

Currently, admission applications are open to the Jhulelal Institute of Technology for the 2025 year. It's easy to use the website, fill out the application and start your learning instantly. When you send in your application early, you are more likely to be accepted. It is advisable to have everything you need ready before starting your final year.

You have to have already graduated from a recognised university to apply for the MBA at Bharath College of Science and Management. About 50% is typically the minimum required, according to college rules. Have your documents and degree on hand before starting the application process

29. $\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)$ $\left(\frac{3-4i}{5+i}\right)$

= $\left[\frac{1+i-2\left(1-4i\right)}{\left(1-4i\right)\left(1+i\right)}\right]$ $\left[ \frac{3-4i}{5+i}\right]$

= $\left(\frac{1+i-2+8i}{1+i-4i-4i^2}\right)$ $\left( \frac{3-4i}{5+i}\right)$

= $\left(\frac{9i-1}{5-3i}\right)$ $\left( \frac{3-4i}{5+i}\right)$

= $\frac{\left(9i-1\right)\left(3-4i\right)}{\left(5-3i\right)\left(5+i\right)}$

= $\frac{27i-36i^2-3+4i}{25+5i-15i-3i^2}$

= . [since, i² = –1]

= $\frac{33+31i}{28-10i}$

= $\frac{33+31i}{28-10i}$ × $\frac{28+10i}{28+10i}$ [multiplying denominator and numerator by 28 + 10i]

= $\frac{924+330i+868i+310i^2}{784-100i^2}$

= $\frac{924+1198i-310}{784+100}$ [since, i² = –1]

= $\frac{614+1198i}{884}$

= $\frac{2\left(307+599i\right)}{884}$

= $\frac{307+599i}{442}$

= $\frac{307}{442}$ + $i\frac{599}{442}$